# Power supply filter caps

#### Wendy

Joined Mar 24, 2008
22,163
Two frequencies, 50Hz and 60Hz.

Two rectification schemes, half wave and full wave.

I need to explain in simple terms how to pick capacitor values for these 4 types of circuits.

Any thoughts or rules of thumb?

This is where I am at on the AAC power supply article.

#### bertus

Joined Apr 5, 2008
20,738
Hello Bill,

Perhaps the attached PDF will help a bit.

I had the remark once that the calculations where not all correct.
unfortunately I can not remember wich one it was.

Greetings,
Bertus

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#### Wendy

Joined Mar 24, 2008
22,163
Thanks, I still have some info Studiot left me a long while back. I need to condense them down into one or two paragraphs, plus math.

#### rjenkins

Joined Nov 6, 2005
1,013
I did this for someone a few days ago as it happens..

Reference: A 1 Farad capacitance with a discharge current of one amp discharges at 1V per second.

For a rough rule of thumb, divide 1 Farad by the ripple frequency and you will get a capacitor value that gives about 1V ripple per amp of load; eg. 10,000uF at 100Hz.

#### MikeML

Joined Oct 2, 2009
5,444
To my dying day, I will remember "Civil Engineers are Queer" (strange!) i.e. C*ΔV = q (Coulombs)

Now i*t = q, so by substitution: i*t = C*ΔV.

Rearrange gives: C = i*t/ΔV.

So if you want a capacitor to deliver 0.75A for 0.0083ms (full-wave rectified 60Hz), say at the input of a LM317, and ΔV must be smaller than 5V to keep the filter minimum voltage from sagging below Vout + VdropOut, then

C = 0.75*0.0083/5 = 0.001245F = 1245uF

#### bloguetronica

Joined Apr 27, 2007
1,372
For full wave rectification, I use the following formula. It is a good rule of thumb:

C = Imax / (2 x Freq x Vripple)

Or you can use this one:

C = Imax / (2 x Freq x Vpeak * ripple_factor)

being ripple factor a value from 0 to 1.

Note that the formula is an approximation, and works better for ripple factors inferior to 0.1 (or for small Vripples when relative to Vpeak).

For the expected DC voltage, check out this guide:
View attachment Rectifier guidelines.pdf

This guide is more precise for AC currents on transformers, though:
View attachment power_xfrmr_Design_guide.pdf

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#### The Electrician

Joined Oct 9, 2007
2,782
Hello Bill,

Perhaps the attached PDF will help a bit.

I had the remark once that the calculations where not all correct.
unfortunately I can not remember wich one it was.

Greetings,
Bertus
There's a strange misconception on page 12-36. Apparently the author thinks that "If the charge time becomes too great, then the capacitor may never reach full charge from the incoming pulses from the rectifier". This is not true, of course.

#### lmartinez

Joined Mar 8, 2009
224
There's a strange misconception on page 12-36. Apparently the author thinks that "If the charge time becomes too great, then the capacitor may never reach full charge from the incoming pulses from the rectifier". This is not true, of course.
How is that so?

#### S_lannan

Joined Jun 20, 2007
246
for a capacitor to instantaneously change voltage you would need infinite power.
In such circuits there is nothing restricting current flow so you can have infinite power to change the voltage to the peak as soon as the diode conducts.

Obviously any small resistance, i.e from the wire from the rectifier to the capacitor would prevent this from occuring in the real world...

#### The Electrician

Joined Oct 9, 2007
2,782
How is that so?
Look at the picture on page 12-35. It shows what happens as the capacitance increases. Now imagine that the filter cap was 1,000,000 microfarads instead of just 1000 microfarads. The slightly downward angle of the blue line would become a nearly horizontal line for 1,000,000 microfarads. It might take a while to reach steaedy state, but once it did, the pulse from the rectifier would keep the cap voltage very near the peak of the sine wave because with a cap of 1,000,000 microfarads the load would discharge it so little that the next rectifier pulse would easily replenish it.

As long as the transformer winding resistance plus the rectifier bulk resistance is substantially less than the load resistance, the DC voltage across the cap will be approximately equal to the sine peak voltage, no matter how large the capacitance.

#### Wendy

Joined Mar 24, 2008
22,163
I've had the same argument with another user on this site, we basically left agreeing to disagree. He was basing his argument off a simulator, mine off of experience and school 30 years ago.

I'm going to cop out some and mention that the transformer will drop its voltage in response to a load, which is true enough, but not give the math behind it. I was impressed with the argument you made about the resistance of the transformer making a difference (resistance, not impedance, which some folks get confused).

#### Wendy

Joined Mar 24, 2008
22,163
To my dying day, I will remember "Civil Engineers are Queer" (strange!) i.e. C*ΔV = q (Coulombs)

Now i*t = q, so by substitution: i*t = C*ΔV.

Rearrange gives: C = i*t/ΔV.

So if you want a capacitor to deliver 0.75A for 0.0083ms (full-wave rectified 60Hz), say at the input of a LM317, and ΔV must be smaller than 5V to keep the filter minimum voltage from sagging below Vout + VdropOut, then

C = 0.75*0.0083/5 = 0.001245F = 1245uF
This one is interesting.

OK, you want NMT 100mv ripple on a 12V power supply at 1 amp, using 60 Hz.

The example I'm going to use is both half wave rectification (t = 16.7ms) and full wave bridge (t = 8.3ms).

For the first:
C = 1A 0.0167sec / 0.1V = 167,000µF

For the second:
C = 1A 0.0083 / 0.1V = 83,000µF

I did this for someone a few days ago as it happens..

Reference: A 1 Farad capacitance with a discharge current of one amp discharges at 1V per second.

For a rough rule of thumb, divide 1 Farad by the ripple frequency and you will get a capacitor value that gives about 1V ripple per amp of load; eg. 10,000uF at 100Hz.
Using the same example,

2st: 1F / 60Hz = 16,000µF, since I want want 1/10 of that then 160,000µF

1nd: Would be double that?

Unless I'm doing a brain fart (not too uncommon) there is a X2 discrepancy, but I'm also assuming a linear relationship, which is not a good idea.

And then there is this:
For full wave rectification, I use the following formula. It is a good rule of thumb:

C = Imax / (2 x Freq x Vripple)

Or you can use this one:

C = Imax / (2 x Freq x Vpeak * ripple_factor)

being ripple factor a value from 0 to 1.

Note that the formula is an approximation, and works better for ripple factors inferior to 0.1 (or for small Vripples when relative to Vpeak).

For the expected DC voltage, check out this guide:
View attachment 16060

This guide is more precise for AC currents on transformers, though:
View attachment 16061
2nd: C = (2 X 60Hz X 0.1V) = 166,667µF
1st: C X 2 ??

I'm seeing an interesting correlation here. I might have to just experiment to pick one, though I like the math in Mike's example.

I also like that we are relatively close in all the numbers, X2 variation really isn't that bad.

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#### bloguetronica

Joined Apr 27, 2007
1,372
...
And then there is this:

2nd: C = (2 X 60Hz X 0.1V) = 166,667µF
...
You forgot to divide. It should be 0.083333F, or 83333uF, according to the formula I use. Thus:

C = 1A / (2 X 60Hz X 0.1V) = 0.083333F = 83333uF (quite near 83000uF).

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#### Wendy

Joined Mar 24, 2008
22,163
I forgot the 1A /, and probably the 2X

C = 1A / (2 X 60Hz X 0.1V) = 83,333µF

This corresponds to Mikes nicely. Looking at it, it is the same formula, just rearranged a little.

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#### The Electrician

Joined Oct 9, 2007
2,782
I've had the same argument with another user on this site, we basically left agreeing to disagree. He was basing his argument off a simulator, mine off of experience and school 30 years ago.
Could you give me a link to that thread? I'd be very interested to read it. I am always interested when simulation fails to give the same result as actual circuitry because it's good to know the limitations of simulation.

#### Wendy

Joined Mar 24, 2008
22,163
I think this was it...

The voltage drop was rather extreme from loaded transformer point of view. I understand transformers loose some voltage regulation, but this was more than that.

I think the simulator wasn't using a line transformer, but something else (such as an impedance matching transformer).

#### lmartinez

Joined Mar 8, 2009
224
Look at the picture on page 12-35. It shows what happens as the capacitance increases. Now imagine that the filter cap was 1,000,000 microfarads instead of just 1000 microfarads. The slightly downward angle of the blue line would become a nearly horizontal line for 1,000,000 microfarads. It might take a while to reach steaedy state, but once it did, the pulse from the rectifier would keep the cap voltage very near the peak of the sine wave because with a cap of 1,000,000 microfarads the load would discharge it so little that the next rectifier pulse would easily replenish it.

As long as the transformer winding resistance plus the rectifier bulk resistance is substantially less than the load resistance, the DC voltage across the cap will be approximately equal to the sine peak voltage, no matter how large the capacitance.
A "Pulse" is not the same as a half wave rectified sinusoidal signal source. Hence, the electrical analysis of the above mentioned document does not applied when a pulse is applied to an RC circuit.

Pule:
http://en.wikipedia.org/wiki/Pulse_(signal_processing)

#### The Electrician

Joined Oct 9, 2007
2,782
A "Pulse" is not the same as a half wave rectified sinusoidal signal source. Hence, the electrical analysis of the above mentioned document does not applied when a pulse is applied to an RC circuit.

Pule:
http://en.wikipedia.org/wiki/Pulse_(signal_processing)
A half-sine such as is emitted by a rectifier may reasonably be called a pulse. That is what the author of the pdf under discussion (http://forum.allaboutcircuits.com/attachment.php?attachmentid=16052&d=1265574755) did; I followed suit. Have a look at page 12-33 of that pdf.

#### Wendy

Joined Mar 24, 2008
22,163
I think I'll run some experiments on this one before I use it as a reality check. The numbers will change but it should be easy enough to plug real world values into it.

#### The Electrician

Joined Oct 9, 2007
2,782
I hooked up a bridge of Shottky rectifiers to a transformer rated at 8 amps @ 24 VAC with a variac supplying the primary so that I could adjust the secondary voltage to about 12 volts. I connected 46,000 μF of filter caps and applied a 1.0 amp DC load. I captured some scope images showing the current pulses from the transformer secondary, and the ripple voltage.

The idea was to get about 100 mV P-P of ripple.

The transformer winding resistance was .127 ohms; that's with the primary resistance reflected to the secondary (and then added in series with the secondary's own resistance). I had a 2 ohm rheostat in series with the secondary before the bridge so that I could adjust the effective transformer winding resistance.

The first image shows the secondary current in purple and the ripple voltage in orange. This is with no additional resistance in series with the secondary; the transformer winding resistance is just the .127 ohms inherent in the transformer. The variac was adjusted to give 12.0 volts DC on the caps.

The second image shows the state of affairs with an additional 2 ohms in series with the secondary. The variac output was increased to maintain 12.0 volts DC on the caps; it had to be increased about 25%.

The third image shows the traces from the first image (shown in gray unfortunately) overlaid with the traces from the second image.

The conduction angle of the rectifiers increased when the 2 ohms was added, and the ripple decreased. Clearly, the winding resistance of the transformer has a substantial effect of the performance of the circuit.

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